WILSON AND LEWIS.- 



RELATIVITY. 



471 



Figure 26. 



Now it is inipossihlo to ro(listril)uto tlie discrete vectors dg over the 

 three iliinensioiuil field from which they were derived, hut it is possihle 

 to rephiee them by a continuous (Hstribution over ii two (hmensional 

 spread in one of the cones. Let us assume tliat the infinitesimal 

 tubes are so chosen that the elements of surface (IS = clq-dr are 

 four-sided fijjures approximately rectangular 

 and that the outer cone is divided into small 

 regions lying between the elements of the 

 cone, a, a', a", . . . (Figure 2(5). In each of 

 these small two dimensional regions we may 

 place the corresponding vector dg. Now 

 any two neighboring lines drawn from a to 

 a' are of equal interval because they lie in a 

 singular plane between two singular lines 

 (see preceding footnote and § 31). The vec- 

 tor dg/dr is therefore determined at each 

 point of the cone independent of the direc- 

 tion of dr. It is a vector representing a 

 kind of density and when all the vectors dg 



are similarly treated, it is continuously distributed over the whole 

 cone. 



The vector dg/dr is a function of the interval ds. Let us determine 

 this relation analytically. Since dS= dqxdr we may write 



dg = (r/qvr/rxl)*l =. [(dq^dT)*'l]\ = U dqdr, ' 



where h is the component of 1 perpendicular to dqxdi; for since dq is 

 perpendicular to dr, (dqxdr)* is a 1 -vector perpendicular to r/q^f/r 

 and of magnitude dqdr. We therefore find dg/dr = Uldq. It remains 

 to determine dq in terms of ds. 



The plane of intersection having been chosen, the two circles are in 

 general eccentric and the distance dc between their centers is the pro- 

 jection of the segment ds upon their plane (Figure 27). If the normal 

 to this plane makes an angle with ds whose hyperbolic tangent is v, 

 then dc = vds/ Vl — r. The two segments cut ofi" by the two circles 

 on dc produced are found as follows. Pass a plane through dc and ds. 

 Ihtn JB iercadilv shown to be 



ds^l 



VI 



' Vl -f ,., and CD = ds Vl -\- v/ Vl — v 



Then the value of dq is readily proved by Euclidean methods to be 



